Question
In: Chemistry
When 200. mL of 0.40 M hydrochloric acid solution is mixed with 3.76 g of aluminum...
When 200. mL of 0.40 M hydrochloric acid solution is mixed with 3.76 g of aluminum metal, how many moles of hydrogen gas would be produced?
6HCl(aq) + 2Al(s) → 2AlCl3(aq) + 3H2(g)
Solutions
queen_honey_blossom answered 1 year agoRelated Solutions
25. Aluminum metal shavings (10.0 g) are placed in 120.0 mL of 6.00 M hydrochloric acid....
25. Aluminum metal shavings (10.0 g) are placed
in 120.0 mL of 6.00 M hydrochloric acid. What is the maximum mass
of hydrogen that can be produced (in grams)?
2Al(s) + 6HCl(aq) → 2AlCl3(aq) +
3H2(g)
a. 11.5 g
b. 0.0505 g
c. 4.76 g
d. 0.727 g
e. 31.8
g
f. 1.42
g
g. 3.67
g
h. 5.13 g
26. What is the
maximum number of grams of ammonia, NH3, which can be
obtained...
25. Aluminum metal shavings (10.0 g) are placed in 120.0 mL of 6.00 M hydrochloric acid....
25. Aluminum metal shavings (10.0 g) are placed
in 120.0 mL of 6.00 M hydrochloric acid. What is the maximum mass
of hydrogen that can be produced (in grams)?
2Al(s) + 6HCl(aq) → 2AlCl3(aq) +
3H2(g)
a. 11.5 g
b. 0.0505 g
c. 4.76 g
d. 0.727 g
e. 31.8
g
f. 1.42
g
g. 3.67
g
h. 5.13 g
26. What is the
maximum number of grams of ammonia, NH3, which can be
obtained...
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please writing hot to solve it .
A 10.0 g piece of pure aluminum is placed in 75.0 mL of 0.54 M hydrochloric...
A 10.0 g piece of pure aluminum is placed in 75.0 mL of 0.54 M
hydrochloric acid at STP condition. They react as follows: 2Al +
6HCl → 3H2(g) + 2AlCl3 Calculate the following:
a. Volume, in liters, of hydrogen gas.
b. Molarity of Al+3. (Assume 75.0 mL solution.)
c. Molarity of Cl–. (Assume 75.0 mL solution.)
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0 mL of HCl, 50 mL of HCl, 100 mL of HCl, 200 mL of HCl, 250 mL
of HCl
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A 22.26-mL sample of hydrochloric acid solution requires 35.00 mL of 0.115 M sodium hydroxide for...
A 22.26-mL sample of hydrochloric acid solution requires 35.00
mL of 0.115 M sodium hydroxide for complete neutralization. What is
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